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What are the little bulges on the tail of an F-18 for?
I noticed on this photo of an F-18 from Wikipedia that the tail of this F-18 Super Hornet has multiple small bulges on it and I was wondering if anyone knows what the purpose of those bulges are?

As visible in the following cutaway (numbers E77 to E80 and F14 - cutaway source), we find:
On the right vertical stabiliser, top down:

Position light
ALQ-214 aft low/high-band receiving antenna
ALR-67 aft receiving antenna
Fuel dump outlet

On the left vertical stabiliser, top down:

ALQ-214 high-band transmitter antenna
ALQ-214 low-band transmitter antenna
ALR-67 aft receiving antenna

The ALQ-214 is a jammer (brochure, pdf) while the ALR-67 is a radar warning system. Note that the ALQ-214 consists also in a forward transmitter antenna (E35) located on the leading edge of the left strake. "Fuel dump outlet" should be self-explanatory

They are antennas
This is the F/A-18A, but mostly the same as your depicted F model.

A Swiss C model:

From https://www.aircraftrecognitionguide.com/mcdonnell-douglas-f-18


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Why is the MIG-21 still used in several militaries even though it is so old?
I was looking through some of the cold was Era jet fighters and noticed that the Mig-21 that was made in the 50s is still in use in places like India and such and was wondering if anyone knows why some countries still use this outdated plane when more modern ones are available?

You might find it hard to believe, but there are even still people driving gasoline-powered cars...

Indian Air Force Mig 21s also recieved consistent avionics and weapon systems upgrade through the years. The Bisons of today are much more advanced than Mig 21s of 1970s.

A Mig-21 is cheap (0.33M) easy to build and has an alright airframe.

Cost vs need. While it isn't the most modern fighter aircraft in the world, that doesn't mean it's useless. Countries continuing to operate it may find that it is still useful for certain types of missions that are important to them, and may not feel a need to upgrade because they don't intend to or stand no chance of competing against a country that is able to field something newer. It can still drop bombs and launch land-attack missiles just fine, even though it would be badly out-classed in a fight against a modern fighter aircraft.
Of the current militaries operating MiG-21 variants, India by far has the largest defense budget. They could buy something new, and in fact they have - their remaining MiG-21s will be removed from service by 2025. None of the other current operators are in the top 40 countries by military expenditure.
So in summary, some countries are buying newer fighters and MiG-21s are among their oldest aircraft, due to be phased out. And as for the rest of the current operators, they don't need the latest and greatest fighter badly enough to justify the extreme acquisition cost.

There's also the issue of availability of replacements to consider.  One big operator in North Korea, who still operate Chinese variants of even older Soviet designs as well.  Few countries would be willing to sell military hardware to  North Korea, and even those that would are going to want payment in currencies that the DPRK is extremely short of (which TBH is all currencies except their own)

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text:
Would a plane similar to a Lockheed Cl-1201 that is capable of deploying jets mid-air be practical in modern day military use?
In the late 1960s Lockheed was reportedly looking into making a massive nuclear-powered jet plane capable of staying in the air for around 41 days and also capable of carrying up to 22 fighter aircraft externally. I was wondering if anyone knows why they chose to not make this plane and whether or not this plane would possibly be useful in modern day air forces.

This plane was never built for a variety of reasons:

The nuclear reactor couldn't be built light enough, and yet safe enough. It would need to survive a collision with a mountain, and needed 20 feet of shielding to protect the passengers. All that added up to weight
The engines worked on the principle of superheating air using heat from the fission reactor. While I haven't been able to find exact details on the engines (probably because they didn't get that far), other engines working on this principle released radiation into the atmosphere, in which case this would have been a flying ecological disaster which would have been forbidden to fly anywhere near population centers
It would have been extremely vulnerable, and easy to destroy with a few missiles. Essentially it would have been a massive target that would be impossible to miss
It would have been insanely expensive to build and operate

In addition to all above there were better, cheaper, safer and more reliable ways to achieve the same mission. None of that has changed.

Point 3  is very debatable.

An addendum for GdD's answer - the actual mission no longer exists.
The fighter carrier concept came from the WWII experience that bombers needed fighter escort to have reasonable survival rate, producing various plans like the Cl1202 to allow a long distance bomber force to bring fighters with them.
The birth of Surface to Air and Air to Ground missiles means that the assumed need for a bomber to actually fly over the target to drop dumb bombs become both impossible and unnecessary.
Air to air refueling also become standard, so 'get aircraft to remote location' simply required using tankers that already existed and broadly useful rather than a dedicated and very expensive new system only able to support a subset of aircraft. Tankers also allow broadly adjustable and flexible strike packages to be sent that will survive some elements being shotdown or having mechanical issues.
While a platform such as this might be usable as a nuclear second strike carrier it would seem inferior to submarines in most cases. Even if the unit cost for one was similar to a submarine they would presumably be much less stealthy due massive IR plume behind it, and therefore could be defeated by upgrading a small number of existing ballistic nuclear missiles to get mid course guidance update via satellite or simply flying tanker supported fighters to shoot them conventionally.
In the deterrence/second strike capability they would also be a major risk to the owning nations territory, as they would need to be continuously airborne producing both operational and crash related nuclear contamination risks. While nuclear submarines also have these problems, they can be made to not leak during operation and their 'crashes' tend to not be in places that upset politicians.

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Why did the CF-105 cease production when it had such promising potential?
I have been reading about the CF-105 recently and noticed that although it was one of the most advanced aircraft of its era it was shut down by the Canadian government after one crashed in a freak accident that was caused by another aircraft and I was wondering if anyone knew if Canada had any better reasons for shutting it down or if it was just  bad move by the government?

The main reason is the same as why many other jetfighters of that era were cancelled, the XB-70 being another emblematic example.
The reason is easy and came (more or less) unexpected, with the Arrow already pointing us into the right direction: "The first Arrow Mk. 1, RL-201, was rolled out to the public on 4 October 1957, the same day as the launch of Sputnik I".
The scientific-political consequences of the Sputnik were quite broad, with the USA for example creating the DARPA and  changing NACA into NASA.
And many of the jetfighters under development in that period were simply cancelled since "it was believed the need for a manned interceptor of very high performance simply did not exist as the Soviets were moving their strategic force to ICBMs". Natural selection.

+1  The UK cancelled the TSR-2 in part because Defence Minister Duncan Sandys' view in the 1957 Defence White Paper was that "the era of manned combat was at an end"


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Why do American Helicopters not have ejection seats or some form of ejection system?
In Russia there is the Kamov Ka-50 which is a helicopter with a ejection seat that blows off its rotors before the pilot ejects. I was wondering why here in America we Don't have some form of similar ejection system to increase the safety of our helicopter pilots.

NOT a duplicate, but a good discussion of the other (i.e. not upwards) ejection options: https://aviation.stackexchange.com/q/43274/7532

Like all things aviation, it's a matter of trade offs, and American manufacturers (and/or military) have obviously decided that the added cost, weight, complexity & maintenance aren't worth the perceived added benefit.

Do not have a source for it, but suspect they can math out the risk of a rotor jettison mechanism missfiring in flight or on the ground killing someone vs a pilot being saved by ejector seat.

Why the reference to America? Even the Russians make VERY limited use of ejector seats in helicopters, and plenty of other countries manufacture helicopters.

@MikeB- I referenced Russia because they are the only ones that I am aware of that do have a helicopter with a ejection seat. Also when I say America I mean the continent and not the country.

Because it was decided on balance that it was not worth it during the design phase of all systems currently in use.
As a starting point, the numbers here suggest that at low level ejection seat survival is around 50%, with helicopters rarely operating at anything other than low level. So at first order approximation fitting an ejector seat will only save 50% of the pilots who use it, and survivors will often suffer serious spinal or limb injuries.
If it is fitted, it needs to be able get the pilot out of the cockpit. For fixed wing aircraft they often either shatter the canopy, or just fire through it but for a gunship helicopter the cockpit is armored and will need some sort of fairly complicated removal system, or a seat that can fire THROUGH the armor.
The rotors also need to be removed. Ideally the main rotor assembly is made very robust, especially for aircraft getting shot at. In this case it needs deliberate weak links with explosive charges fitted, probably up in blades where signals need to get from fixed airframe into rotating shafts. Loosing a blade because weak link or associated charge/wiring was damaged is not desirable.
It is really important a partial activation is avoided (seat fires while rotors or canopy still present, or rotor fires but seat does not get message so aborts). This is also relevant for fixed wing aircraft but all the needed systems are either in the seat or adjacent canopy area (and a canopy failure may not prevent system operating, just make injury more likely). If the pilot is healthy enough to trigger an eject there are good odds the seat they are sitting in is also OK.
For a helicopter system the most likely reason to be ejecting is damage to rotor or engine. So we need to design a system that gets the needed signals to the rotor assembly, and back to confirm it triggered despite damage to the area. This presumably involves multiple redundant links and some sort of voting scheme that will be complex to design and maintain. Making it wireless would solve the damage to wiring problem but allow a remote hack to crash all your helicopters!
Then we have the inverse situation, helicopter with this system has a minor accident and hits the ground. The ejector system will need a substantial number of pyrotechnics, and may have systems to retain the pilot during ejection, this may hinder pilot escaping from the crash and mean that if pilot does have issues escaping they are sitting on top of a large amount of possibly damaged explosives. Canopy jettison or seat fire while crew or rescue team partially inside cockpit is also highly dangerous.
These issues also apply to servicing, where all explosive charges need to be safed during work and then re-armed prior to flight, especially those capable of clearing an armoured canopy or dismantling the rotor assembly.
Currently the focus seems to be allowing the pilot and crew to survive a crash, which has the advantage of generally also working for hard landings and other intermediate incidents where an ejection seat is very much an all or nothing deal that guarantees loss of aircraft and most likely permanently injure the crew.
A final darker aspect is survival post activation, given that helicopters are often operating very close or on top of hostile forces. Going down inside an armoured cockpit from 50 meters may be safer than hanging under a parachute at 500 meters against a force that considers bailed out pilots eligible targets.
None of this means ejector seats are will never exist for helicopters other than ka-50, just that not fitting them will often make more sense. An example case where one might show up is a human sized multi rotor, where firing up is much easier and auto rotation to safe landing post engine failure much harder to engineer.


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How did the engines on the ME-262 work?
I have been looking into the ME-262 jet engines recently but cannot quite understand how they worked and was wondering if anyone could explain it to me in a simplified way.

It had a "standard" turbojet: do you want to know how a turbojet works? Or something else?

Please edit your question to specify what it is about the Me-262 engines you don't understand, beyond what is explained in the answers to this question. Otherwise, this question may be closed for not being clear as to what is being asked.

The Me-262 engine sucked air into the front of the engine with a series of rotating fans, compressing it. Fuel was then sprayed into the flow of compressed air and ignited, causing the fuel to burn and the surrounding air inside the engine to heat up and expand. Those expanding gases had to accelerate on their way out the exhaust duct and another fan inside that duct captured some of the kinetic energy in the exhaust flow and used it to spin the compressor fans. The leftover momentum in the departing exhaust downstream of the exhaust fan was greater than the momentum of the air entering the engine and the difference in momentum of a parcel of air passing through the engine per unit time has units of force, which is the thrust of the engine.


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How does a VTOL mode work on a Harrier Jet?
I have heard that Harriers use downward jet exhaust to do VTOL Maneuvers and I was wondering if someone could explain how that works?

The Harrier achieves vertical lift by having four thrust-vectoring nozzles positioned on the sides of the fuselage, coupled with one turbofan engine. Here's how the nozzles look on the actual fuselage:

Image via Alan Wilson on Wikimedia Commons, CC-BY-SA
And here's a diagrammatic view of how the engine is coupled to these nozzles:

Public domain image via Fama Clamosa on Wikimedia Commons
When vertical lift is desired, the pilot moves a lever next to the throttle in the cockpit and sets the desired angle (which is itself a function of speed and phase of flight). The pilot may additionally dump water from a reserve tank into the engine exhaust to provide extra lift.
When configured for vertical flight, the Harrier faces a number of challenges:

Disrupted airflow into the engine intake: not only is less air being rammed into the engine as would be the case during forward flight, but the air is now moving perpendicular to the engine's axis.

Extremely limited control authority on all three axes: with almost no airspeed, there's no real yaw, pitch, or roll control, and the vectored thrust nozzles don't include differential thrust (as far as I know).
Newer platforms like the F-35B incorporate wingtip nozzles for improved roll authority and fine, automatic rear nozzle vectoring to provide a safer, more controllable experience, and are further augmented by computers to relieve the pilot from managing all these extra features.

High fuel burn: the Harrier only provides enough vertical thrust to lift itself, a little bit of its fuel, and only the lightest armament configuration in most conditions, typically with the throttle nearly at full power. Water can be dumped into the exhaust to provide some extra thrust, but this reserve usually lasts 2-3 minutes. This is all caused by the engines having to provide all the lift, unassisted by the wings.

Little or no computer assistance, especially when compared to modern counterparts like the F-35B. Pilot error in the Harrier's VTOL regime had the potential of very quickly becoming catastrophic, between the dulled input responsiveness and high pilot workload.

In general (though definitely not always), Harriers would perform vertical landings on underway ships rather than at airfields. The 15-30kn of forward motion was enough to get the wings to generate some lift, mitigating a handful of the issues I listed above.
For more information, please see the Wikipedia page on the Harrier family.


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Could you encode data in a Black Box ping?
Black boxes put out a sonar ping at about 32kHz. From what I've read, depending on sea conditions the ping will be audible at the surface out to about 16 km, leaving a search area of around 1024 square km.
By adding a little complexity to the ping circuit, you could encode within it the last received GPS position of the box, before it hits the water. With average sea depth and average sink angles of less than 45° that information would cut the search area down to about 10 square km.
Is there some reason this tech is not used, or is it still in the rigorous testing pipeline these devices must go through?

under sea currents would still cause a large drift, also directional detectors (and just going a mile in one direction and detecting differences in the ping strength) will do more than an inaccurate GPS lock

What is the black box's "last received GPS position"? The black box is inside the plane: it can't receive GPS because it doesn't have line-of-sight to anything, including the constellation of GPS satellites.

Current underwater locator beacons (ULBs) have a detection range of only a few km. 16 km would be under exceptionally favourable conditions. So in most cases the GPS data would not be of any additional help. It might tell you where the aircraft was before it hit the water, not exactly where the debris ended up after falling through several Km of water.
An advantage of current ULBs is their extreme simplicity and small size. There's no need for external data-feed and power connections. There are no issues of openings in the casing for waterproof connectors that also need to withstand the g-forces and pressures expected to be endured by a ULB. They are easy to test regularly and relatively easy to maintain.

From what I have read, current proposals for improvements to ULBs mostly focus on

increase duration from 30 days to 90 days
increase range by changing frequency from 37.5 kHz to ~10KHz

There are trade-offs here - longer duration means bigger batteries. changing the frequency may make it harder to distinguish from other sources of sound in the ocean.

An interesting 2011 dissertation describes the difficulties involved  and discusses modulating the beacon signal to include data.

and GPS doesn't work under water!

@rbp : yes, but the GPS data of the last known location could be used.

Even if one started a search with many vessels whose equipment can detect nothing more than that a beacon exists somewhere within 16km, a ship that detected a beacon's signal could summon one or more ships with fancier equipment which would be able to home in on the signal much more accurately.  I don't think that encoding a plane's last above-ground location in the ping data would assist significantly with locating the beacon.
Having the black box "blindly" encode any significant amount of information in its pings would add greatly to the power requirements; there might be some benefit to having a black listen occasionally (the pings would indicate when) for an outside request for data, and then transmit some of its data in response to such requests.  This could be helpful in cases where the box could be located, but was in a place that salvage equipment could not reach.  Only a limited amount of data could be transmitted, however; depending upon how far away it needed to be received, transmitting even a minute's worth of audio might use more energy than a month's worth of pings.  I don't know that the number of cases where a box is detected, and holds data, but cannot be physically retrieved, would be sufficient to justify the extra cost and complexity of ultrasonic data communication.  It would probably cheaper to add redundant recorders to the aircraft in places that would likely end up in different parts of the debris field, so as to maximize the likelihood that at least one would be reachable by salvage craft.

Main reason those signals can be distinguished for what they are is because they have a specific signature. If the signals were to change all the time because you mix in extra information, they'd be impossible to recognise and thus useless.
Combined with the short range of the signal (inevitable consequence of the need to be low power, anything more would mean heavier equipment and/or shorter lifespan), there's just no benefit.

That doesn't follow. For example, radio receivers can only pick up radio waves that "have a specific signature" but those radio waves can be modulated to encode data and, obviously, this can be done without breaking the "specific signature".

@DavidRicherby such would mean replacing every single receiver out there, just like you now need AM, FM, LW, etc. etc. receivers for every single type of radio signal. And all for nothing. No, modulation is not the answer, it only adds complexity for no reason whatsoever (but then, that's politics, doing things for no reason whatsoever so I'm sure you'll get it mandated by the FAA if you just demand "they do something NOW!").

You would only need a new receiver if you actually want to decode the data. A simple receiver could still detect the fact that there is a signal, just as it would now.

@fooot Indeed: receiver records signal, sends to HQ, HQ decodes, HQ says "The message was xyz". I'm not saying this is a good idea, just that jwentings objections to is aren't well-founded.

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Why do these safety instructions have a ruler on them?
I was on a Wizz Air flight the other day and the safety instructions leaflet had a CM ruler running down the side of it.
What is the purpose of this?

This looks like just a courtesy tool on the safety card, but doesn't actually have a requirement or purpose. Previous versions of their safety card did not have this courtesy tool.

(Image Source: aviationfreaks.freeforums.org)

I'm quite curious: Is there any other airlines that put such courtesy tools on their safety cards?

The rulers are there for measurement of anything you wish to measure in flight (which mostly includes measuring the sizes of the baggage of the consumer if there is a doubt about the size). Such rulers come in handy as big bags which are above the foretold measurements are quite common, so the flight attendants leave it up to the consumer to measure their own bags and not cause any hassle to the airlines.

Wizzair, as well as many other low-cost carriers, actually measure the carry-on luggage before entry to the aircraft if they doubt it fits, mainly because they charge a horrifically high surcharge if you need to check in your luggage last-minute due to size. The ruler on board on the safety card would rather not be fetched to do that.

Usually, your luggage size is measured ONLY if cannot fit in the given places, and that is when the ruler would be used

Ryanair has a cardboard luggage checker that they simply put over the luggage in the gate area. To check the luggage inside the aircraft would cause unwarranted delay.

@VivianjeetSinghSudan If it doesn't fit, that seems conclusive to me. It's hard to see how a ruler would help there.

@DeltaLima, because some airlines have a rule which states that the passenger can still carry a cabin luggage bag if it does not fit but is smaller than a given dimension, after paying a nominal fee

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Is the A321 XLR really necessary to fly MAD-BOS?
I am a bit puzzled by the routes that the A321 XLR is (about to be) operating.
For example, MAD-BOS is barely 3000 NM, meaning even a 321 not even LR would be able to make it.
Unless I missed something, not a single of the announced routes would require the 4700NM range of the XLR, an LR would be able to cover them all.
Any thoughts on this?
Thanks!

Note that the range quoted by the manufacturer is without wind (and there is usually significant headwind over the Atlantic due to the jetstream) and does not include additional reserves that may be needed for EDTO (Extended Diversion Time Operations) flights over oceans.

Any chance you own an electric car?  Stuff about spec range vs adverse conditions range will become more experience-born, including, relevantly, wind.

The "maximum range" given by the manufacturer is not the same as the maximum range of a typical airline flight. An airline flight has to consider additional factors that are not included in the basic payload/range calculations:
Wind - If there is a headwind, the range will be reduced.
Temperature - Above standard (ISA) conditions will reduce performance and range.
Regulations -
Having "just enough" fuel to reach your destination is not a good idea in practice. An airline flight will also add contingency fuel to account for any unexpected conditions, alternate fuel to allow a missed approach at the destination and flying to an alternate airport, and then final reserve fuel. There also needs to be enough fuel to reach a diversion airport in the case of engine failure. MAD-BOS is an oceanic flight, meaning that diversion airport may be hours away, and fuel consumption in this situation becomes higher than normal due to the failed engine and lower altitude required to fly on the remaining engine. All of these factors reduce the actual range capability.
If you look at the generic payload/range diagrams from Airbus, the A321LR (97t MTOW with 2 or 3 ACT) just barely achieves a maximum range of 3,000 NM at max payload. The extra fuel required for the reasons above would require reducing the payload (and therefore revenue) for the flight. The extra range of the A321XLR allows full payload capability while carrying the required fuel.
So to answer the question: no, the A321XLR is not necessary for a route like MAD-BOS, but it is much better suited than the other A321 variants.

Source

If the airline operated an all-business-class flight, then an A321LR would fit the bill nicely.


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Can an APU be designed to also provide emergency thrust in the event of a double engine failure?
Commercial airliners seem to be moving towards twin jets with incredibly powerful 100,000+ lb thrust engines becoming available for use.  Would it be possible to design the APU to provide enough thrust to push an aircraft to a nearby runway in the event they
both fail as a safety measure?

This is more of a philosophical argument: no, it is not possible, because then the APU would not be an APU, it would be an engine.

It'd be a mighty big APU.  Might as well call it an engine.

No, it would not be possible with current technology.
To scale the APU to an engine that provides thrust, you would basically have to turn it into another engine, with significant weight penalty. The position of the APU inlet and outlet would have to be re-designed, not to mention mechanisms that allow it to produce variable thrust. In short, it would be turning the plane into a tri-jet.
Furthermore, the chance of a double independent engine failure are slim. If a condition would lead to both engines fail, it would very likely lead to failure of any remaining engines. Converting the APU into a thrust producing device would be a complex engineering exercise with little to no benefit in risk reduction.
Perhaps decades later, we can produce a powerful jet engine using lightweight materials.

tj1000 provided the correct answer (though it did not receive proper recognition). The APU is a turboshaft engine - it's designed to harness as much of the energy from the exhaust, and pump it into the shaft. While turbofan engines work in a similar fashion, pumping energy into the fan, the APU does not have a fan. Also turboshaft engines usually have centrifugal compressors as opposed to the axial compressors on the turbofans.

Keep in mind that the APU is a turboshaft engine, not connected to a prop but a generator. Therefore, any exhaust exiting the APU will have lost most of its thrust to the shaft turbines, or to bleed air used to power any aircraft systems that would normally be powered by main engine bleed air.
Sure, you could enlarge the APU to the point where it could supply thrust, but... it would be very inefficient as an APU (gas turbines don't run at low power settings very efficiently), it would add substantially to both weight and drag when it wasn't in use, and it would be essentially another main engine.
The whole point of an APU is to supply power and bleed air efficiently to an aircraft when the main engines aren't running... when it's parked on the ground. Typically, on an airliner, the APU also supplies bleed air to start the main engines if ground based bleed air isn't available. And typically the APU is small so it can be started by the aircraft's battery.
The keyword being efficient... for which it must be as small and light as possible, and consume as little fuel as possible. Given the very high reliability in today's gas turbines, an APU that can also supply sufficient thrust to keep the aircraft flying wouldn't be necessary, and would make for a fuel hungry aircraft.

I would recommend adding that turboshafts tend to have centrifugal compressors as opposed to axial compressors in turbofans.

Quote from Jane's Aero Engines 2017-2018:

APU: auxiliary power unit. An easily started gas turbine, usually forming a self-contained package inside the aircraft, which, either on the ground or following main-engine failure, can suply some of the following: shaft power, hydraulic power, electrical power, high-pressure aire, and (in a few recent cases, mounted in the tail) emergency propulsive thrust.

@ymb1 The text I quoted is from a preface devoted to terms definition. That book is in my local library so today I search if there is any jet engine that is propulsive APU comprised


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What produces thrust along the line of flight in a glider?
After reviewing discussions and vector diagrams of gliders in flight, the vertical lift component and the vertical drag components seem to produce a steady state, 0 acceleration balance with gravity. The glider is on a straight line flight towards the ground. Would the forward motion of the glider be accounted for by combining the horizontal lift vector and gravity? Would any other forces be involved?

See also: How can an airplane keep going forward if it has no thrust?

Comments are not for extended discussion; this conversation has been moved to chat.

V is the airspeed. U and w are the horizontal and vertical components of V. The rest of the vectors are forces; W = weight, K = aerodynamic force, L & D are the orthogonal components of K. Those are all the forces involved in a glide.

I was hoping someone would say V is a combination of velocity down and velocity forward (created by horizontal component of lift vector).  Interesting that, even though acceleration forces are in balance.  Can we have a "relative velocity" direction based on falling and moving forward at a constant speed?

Better yet, what generates "U"?

The projection of the weight vector is the 'thrust' that opposes the projection of the aerodynamic force (drag). As those forces are in equilibrium, the motion is uniform, with a constant speed and zero acceleration.

Airspeed is not a force. V would be the trust component. The question is what is providing the thrust component that is opposing drag.

Force U is 90 degrees to gravity there for can not be created by it.  The horizontal component of lift is not drawn (why?).  Gravity cannot be moving the plane forward any more than gasoline can provide thrust.  Is not the wing the "engine" of lift and thrust, as in birds?

Although I do agree that vertical drag plus vertical lift create a steady state descent.

@Robert DiGiovanni, is there a specific problem you are trying to resolve here?  I am having a hard time understanding your confusion over such a simple concept as a glider.

Michael, you need to speak to the person who made the diagram.  I happen to be in full agreement with quiet flyer that it involves lift, drag, and weight vectors.  If you have ever experienced building and throwing a free flight glider, and enjoyed its flight, you might understand my nit picking.  Now start with a parachute (or hang glider), shift your weight in one direction what happens?  Why?  How does airfoil achieve proper angle of attack to function as a lifting device?  How is straight line flight achieved and maintained?  This is the core of understanding gliding.  It interests me.

Now, the question is "What creates force U?"

"I do not see the purpose of the line connecting K and L, other than to make it "seem" right."-- that is to show that the drag vector can be redrawn in that position, to show the K is the resultant (vector sum) of L and D.

"Now, the question is "What creates force U?""  -- U is not a force.  U and w are components of the airspeed vector V.

"The horizontal component of lift is not drawn (why?)"-- it's clear from the diagram that Lift, Drag, and Weight can be arranged into a closed vector triangle.  That's all we need to know about what forces make gliding flight possible.

Now we get to airspeed vector, which is created by drag bringing acceleration forces into balance.

K is not necessarily equal to W.

@TomMcW W has two components, Wn and Wt.  Wt is in the same direction as V and opposite to D.

"What produces Thrust along the line of flight in a glider?"  Nothing.  There is no Thrust involved in gliding flight.  Lift, Drag, and Weight form a closed vector triangle, meaning net force is zero, meaning no acceleration or deceleration is taking place.  The Weight vector can be resolved into components perpendicular and parallel to the Drag vector, but noting that Weight has a component that is parallel to Drag is not the same as saying that a Thrust force exists.
See these vector diagrams:
The balance of forces in gliding flight can be represented by a closed vector triangle of Lift, Drag, and Weight, with right angle between the Lift and Drag vectors.  If we wish, we can also show the airspeed vector, and we can break it down into its horizontal and vertical components.  The vertical component of the airspeed vector is simply the sink rate.

All three of the above vector diagrams represent the exact same physical situation.  We've represented the aerodynamic forces in black, and the gravitational force-- the weight vector-- in green.  The left-hand and right-hand diagrams are easier to understand than the middle diagram, because the force vectors are arranged into a closed polygon (a triangle in this case), which shows us instantly that the net force is zero.  On the right-hand diagram, we've also included the airspeed vector, and its horizontal and vertical components.  Note that K is the glide angle, and note that on the right-hand diagram, the triangle formed by the force vectors and the triangle formed by the airspeed vectors are similar triangles.  The right-hand diagram makes it immediately obvious why the ratio of Lift to Drag, is also equal to the glide ratio through still air-- i.e.the ratio of horizontal speed to vertical speed (sink rate).  Note that for high L/D ratios, Lift is very nearly as large as Weight, and horizontal speed is very nearly as large as airspeed, so for most practical purposes we can use airspeed / sink rate, glide ratio, and L/D interchangeably.
Note that if we wish, we can break the Weight vector down into two components, one of which is exactly equal in size and opposite in direction to the Lift vector, and one of which is exactly equal in size and opposite in direction to the Drag vector.  The latter can be viewed as being somewhat like Thrust, but it's not strictly correct to label it as such.
The lift vector is proportional to the lift coefficient times the airspeed squared.  The drag vector is proportional to the drag coefficient times the airspeed squared.  The ratio between the magnitude of the Lift and Drag vectors is identical to the ratio between the lift and drag coefficients.  For a given fixed aircraft shape, a change in the lift coefficient and/ or drag coefficient represents a change in angle-of-attack.  Every possible angle-of-attack of the wing is associated with a specific lift coefficient, a specific drag coefficient, and a specific Lift : Drag ratio.  This is true regardless of whether the aircraft is heavy or light-- a change in weight is expressed as a change in the airspeed associated with any given angle-of-attack, but no change in the lift coefficient, drag coefficient, or Lift: Drag ratio associated with any given angle-of-attack.
It is crucial to understand that the above vector diagrams will be valid for flight at some given particular angle-of-attack of the wing REGARDLESS of whether the airmass is rising, sinking, or neither.  A glider doesn't feel any difference in the direction and magnitude of the forces acting on it when rising in a powerful mountain wave, than when gliding in still air.
Note also that we're only dealing with the wings-level case.  In circling flight, the forces still look the same in a side view, but airspeed and lift and drag are all higher than in wings-level flight at the same angle-of-attack, and the glide ratio is decreased, and from a head-on view we'd see that the part of the lift vector has been diverted to the side.
Back to the simple case of wings-level flight--
What is "powering" the glider?
Work is proportional to force times distance, and power is proportional to force times velocity.  In the airmass reference frame, the glider's direction of travel is identical to the direction the airspeed vector is pointing.  The only force doing work on the aircraft in this direction is the component of the weight vector that points in the same direction as the airspeed vector.  We can break the Weight vector down into one component that is equal and opposite to the Lift vector, and one component that is equal and opposite to the Drag vector.  The latter component is what is "powering" the glider, at least as viewed from the standpoint of the airmass reference frame.  The latter component of the Weight vector will be equal to Weight * cosine (glide angle).  So the Power exerted by this component of the Weight vector is equal to (Weight * cosine (glide angle)) * airspeed).  Now, it just so happens that the sink rate is equal to (airspeed * cosine (glide angle)).  So we can say that the power exerted by gravity = (Weight * cosine (glide angle)) * airspeed) = (Weight * (cosine (glide angle)) * airspeed) = Weight * sink rate.
As viewed from the ground reference frame, the glider's direction of travel is different when the glider is in rising airmass than when the glider is in still air, so the above formulae for work and energy will only be valid in still air.  From the ground reference frame, the glider's direction of travel is sometimes horizontal or climbing, in which cases the Lift vector can be viewed as having a component that is doing work on the glider.  But as noted earlier, the direction and magnitude of the Lift, Drag, and Weight vectors remains the same regardless of what reference frame we are using for our velocity vectors.
(Note-- this answer is focused on linear steady-state gliding flight.  It is not intended to address a case where the glider is accelerating.  Note that a change in airspeed will affect the magnitude of all the aerodynamic force vectors, and note that any imbalance between L and the opposing component of the W vector will make the flight path curve up or down.  The topic of how, starting with some imbalance between L D and W, a glider transitions back to a steady-state case where L, D, and W are again in balance (in the absence of corrective control inputs from the pilot, typically via a series of gradually decreasing "phugoid" oscillations involving continual variations in glide or climb angle, vertical speed, airspeed, pitch attitude, and to some extent angle-of-attack), is a fascinating one but beyond the intended scope of this answer.  Likewise the subject of loops and other aerobatic maneuvers, or any other form of accelerated flight.)
See related answers to related questions:
An airplane has an engine that pushes its flight. What force pushes a glider to fly?
"'Gravitational' power vs. engine power" -- https://aviation.stackexchange.com/a/56040/34686
"Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind?" -- https://aviation.stackexchange.com/a/56259/34686
"Does lift equal weight in a climb?" --
https://aviation.stackexchange.com/a/56476/34686

@RobertDiGiovanni weight can only go straight down and cannot be resolved into vectors. But you asked what propels it along it's flight path. Gravity can be divided into vectors wrt an angled flight path.   BTW, if you got what you came for you should accept one of the answers.

"weight can only go straight down and cannot be resolved into vectors. "- sure it can. The force vector called Weight (= mass * grav. constant) can be resolved into an infinite number of pairs of two other vectors, each with a 90-degree angle between them. For the purposes of this diagram someone decided to resolve Weight into two vectors acting parallel and perp. to the flight path. Personally I think it makes a better diagram if you just draw a closed vector triangle involving Weight, Lift, and Drag and be done with it; forget about resolving the forces into pairs of orthogonal components.

This answer would be improved by noting somewhere near the beginning, that lift is less than weight (lift = weight * cosine glide angle), as shown by the diagrams.

@quietflyer You have a problem conceptually.  DWL forms a triangle, but it can only be said that they are balanced.  That is, "thrust" is balanced with D.  So it was the glider moving at a constant speed.  Of course, DWL can also be unbalanced.  When the "thrust" is greater than D, the speed of the glider will increase.

@enbinzheng -- DWL either form a closed vector triangle, or they don't.  I'm addressing steady-state constant speed linear flight, where they do form a closed vector triangle.  So your comment is really off-topic from my answer.  Talking about the dynamics of how the glider comes back balance starting from a given situation where there is not a closed vector triangle is an interesting and complex topic, but off-topic from my answer.

@enbinzheng -- join me in a chat room I'll now create if you want to explore this further; trying to do it here will inevitably get all these comments moved because comments are not intended for extended dialog.

@enbinzheng and others -- https://chat.stackexchange.com/rooms/105834/force-vectors-in-gliding-flight

@enbinzheng -- now see disclaimer at end of answer.  Does that fully address your concerns?

future edit-- link to another answer to address the more general case where we don't have linear steady-state flight and L D and W don't form a closed vector triangle.

future edit-- address the issue of considering what force acts along the flight path as seen from the ground and link to other answer that adopts that viewpoint.

In a steady state with no acceleration nothing needs to produce a forward force. Inertia indicates that all forces are equal. This is true whether climbing, descending or maintaining constant altitude.
That given, there will always be a drag force trying to slow the glider along its flight path. To remain at a steady airspeed it is only this drag force that must be overcome. Since a glider produces no thrust, drag must be compensated by angling the flight path downward causing a portion of gravity to counter the drag force. If you divide gravity into x and y coordinates referenced to the flight path it is the portion that is parallel to the flight path that is countering the drag.
This is all presuming no updraft.

Even in an updraft the forces are exactly the same, and the velocity vector drawn RELATIVE TO THE AIRMASS REFERENCE FRAME is exactly the same.  If you happen to be standing in a rising elevator as you look at the diagram, you'll know what I mean.

@quietflyer Yeah, but an updraft complicates things. The flight path is no longer the same angle wrt gravity and it provides an additional aerodynamic force. I was just keeping it simple

Actually the flight path with respect to the airmass is at the same angle to the horizon in updraft, downdraft, and still air, and  the aerodynamic forces are identical in all three situations.

@quietflyer https://engineering.stackexchange.com/q/26575/19873This is a human-powered aircraft I have considered. Do you think it is possible?  Welcome to discuss.

What creates force U?  We know the glider is descending vertically at a constant velocity. We know the wing is pitched down at an angle. How does this translate into a horizontal velocity component? As the "wind" from the vertical descent strikes bottom of the wing, some of the energy will move it sideways!
Notice a perfectly circular parachute will drop straight down.  With a sinking glider, the horizontal stabilizer will push the nose down, creating an angled surface to the wind.  This is how gravity (the fuel) uses the wing (the engine) to create forward thrust (Force U).
In the steady state, the flight path would be the result of forward and vertical velocities (U and w).
Comments would be most welcome.
AEROdynamics. Pencil and paper ready. Draw a straight line along the bottom of the paper (ground). Draw a horizontal line across the middle of the paper (horizontal flight level). Take a point in the center of the flight line. This is your CG. Draw a vector straight down towards the ground. This is your weight vector W. Draw a vector upwards and to the right of CG (of equal length). This is your total lift Lt. Draw (head to tail) the horizontal and vertical components of the lift vector: Lv and Lh.
What do we have? Forces out of balance. The vertical lift vector Lv is smaller than W. Lh is unopposed. There for, the point must accelerate down and to the side. What will stop the acceleration, drag from velocity. I realize this is long hand compared with Quiet's analysis, but it solves by (more realisticly) dividing drag into vertical and horizontal components while keeping gravity vector down and accounting for direction of flight V.
Now draw in any glider you like.

One way to get a "feel" of how strong the pitch angle influence motion is to try to push a knife, flat-side-down, through butter (or jelly). If the knife is slightly angled you will find that the butter forces the knife to move forward instead of down

U is not a force.  It is a component of velocity.  V is the net velocity and U and w are components.  K is the net aerodynamic force and L and D are the only components of that force. Remember, to create a steady motion (constant velocity), there is NO requirement for a net force to be acting along the direction of motion.  Rather, there is a requirement that the net force MUST be zero.  If what you really asking is "How is drag overcome-- how can the net force along the flight path be zero if there is drag", the answer is Gravity. L, D,and Weight form a closed vector triangle. Zero net force.

It would be a misconception to think that the LIft vector somehow contains a component that acts opposite to drag. The references to the wing creating forward thrust are inaccurate-- yes the wing's lift has a forward component relative to the horizon, but at the same time the wing's lift vector is completely perpendicular to the drag vector.

Don't feel bad, even NASA can't get it right.  There is no way the L, D, and W vectors shown in this diagram https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html can be arranged into a closed vector triangle.  The proportions are all wrong.  Why doesn't a few minutes of googling the web turn up a nice diagram where  L, D, and W are actually arranged in the form of a closed triangle?  Yet I am having trouble finding such a thing.

I hope a good L, D, W diagram can be found.  In reality, gravity does run the show.  The CG and neutral point, horizontally off set, cause the nose to pitch down once the plane starts falling.  The velocity is controlled by the elevator trim.  The wing pulls the plane sideways anyway you look at it.  (best from nose straight down drop).  The vectors and L,D,W triangle are the result, not the cause, of the glide condition.  Thanks for a great discussion.  L,D,W will be checked out.

"(best from nose straight down drop)"-- if you want to talk about the complete dynamics involved in dropping a glider with the nose straight down and watching the glider eventually transition into a steady stable glide, it will get much more complicated than we have touched on so far.The pitch "phugoid" will play an important role. Now we're talking about stability and control-- that's a whole different subject from the basic balance of forces in a steady glide. There's no objective criteria by which to say "The vectors and L,D,W triangle are the result, not the cause, of the glide condition."

W never changes.  L and D and V are determined by elevator trim and will stabilize as the glider does.  As far as being "phugoid", my gliders don't do that.  I would imagine setting weight too far forward with too strong an elevator trim might induce it.  But you are right, this is a stability and control issue.

@quietflyer According to your logic, D is balanced by the component of gravity.  Right?  Otherwise, WLD will not have zero net force.

@enbinzheng I will answer you here https://chat.stackexchange.com/rooms/105834/force-vectors-in-gliding-flight

@RobertDiGiovanni Some people think that there are only $L$, $D$, and $W$ on the glider. This is not entirely correct.  Simply put, there are only two forces on the glider: one is aerodynamic $F$, and the other is gravity $W$.  $L$ is only the component of $F$ in a direction perpendicular to $V$; $D$ is only the component of $F$ in a direction parallel to $V$.  Some people think that $Wn$ does not exist, which is also incorrect.  $Wn$ is the component of $W$ in a direction perpendicular to $V$; $Wt$ is the component of $W$ in a direction parallel to $V$.

@enbin zheng I enjoy seeing other points of view here, and your writings are helpful.  Starting with the block on an inclined plane: only 2 forces, W and D.  We put a wheel on the block, D becomes wheel bearing friction F.  The component that "pulls" the block will be drawn forward of the block CG, but Wn can also be imagined from the block reference.  Note as we tilt the block further forward, Wt gets bigger, until it overcomes D.

@RobertDiGiovanni I mean it's not L but Wt that pushes the glider forward.

@enbin zheng it does! But only if vertical drag and vertical lift components = W.  When drawn from the aircraft perspective W is "forward", when drawn from the earth perspective, L is "forward" (plane is pointing down).  Really 2 ways of saying the same thing. The glide path is the same as the ramp the block rests on, supported by aerodynamic forces.

@RobertDiGiovanni Instead of looking at L from the ground, L's component is the force pushing the glider.  Note that L has no component in the direction of V.  When V is pointing down, it is always Wt that pushes the glider.

@enbin zheng agreed

The weight $W$ of the glider has two components, $Wt$ and Wn. $Wt$ is in the same direction as $V$ and $Wn$ is perpendicular to $V$.  Aerodynamic $F$ also has two components $L$ and $D$, where $L$ is perpendicular to $V$ and $D$ is parallel to $V$.
When $L = Wn$:
$Wt> D$, the linear acceleration of the glider is positive; $Wt <D$, the linear acceleration of the glider is negative; $Wt = D$, the linear acceleration of the glider is zero.
Some people think that there are only $L$, $D$, and $W$ on the glider. This is not entirely correct.  Simply put, there are only two forces on the glider: one is aerodynamic $F$, and the other is gravity $W$.  $L$ is only the component of $F$ in a direction perpendicular to $V$; $D$ is only the component of $F$ in a direction parallel to $V$.  Some people think that $Wn$ does not exist, which is also incorrect.  $Wn$ is the component of $W$ in a direction perpendicular to $V$; $Wt$ is the component of $W$ in a direction parallel to $V$.

What pushes the block downward?
 $V$ is the speed at which the block moves downward.  The weight of the block has two components: Wn perpendicular to $V$ and Wt parallel to $V$.  $L$ is the reaction force of the slope (inclined plane) on the block.  It is not the $L$ push block that moves downward, because $L$ has no component in the direction of $V$.  Wt is the force that pushes the block downward.

Comments are not for extended discussion; this conversation has been moved to chat.

@enbinzheng while I think this is a good treatment of the general case where the glider is not necessarily constrained to be not accelerating, I do note that the original question said that the glider was not accelerating.  But I hope you to decide to just leave this up as a good treatment of the general case rather than modify it in which case it would likely up just duplicating some of the other answers.

It might be easier for you to understand this from an energy balance standpoint, as follows: The vertical descent rate in feet per second of a glider times its weight yields the energy release rate (foot-pounds per second) being dissipated by drag forces as it descends.  Dividing the energy release rate by 550 yields the horsepower required to maintain the glider in its glide. This energy is being dissipated by a drag force applied to the airframe; that force times the airspeed yields the drag horsepower. In a steady-state glide, the descent horsepower equals the drag horsepower.

Ok, for (academic) fun, we have gotten from fuel (gravity) to energy release (engine), what is the propeller ?  What gets the glider moving forward?  Very basic, but important, as this is the "incipient glide" Once the aerodynamic (from falling) forces push the nose down and airflow starts over the wing (propeller). Gravity does not do this.  Airflow does.  Diagram could be better.

But I do like the concept of steady state drag horsepower, as it could be used to determine thrust requirement for powered flight.  I had imagined Weight x Inverse of glide ratio.  Applied it to can a 747 fly on one engine and it was close!

Robert don't you remember that this was recently explored in detail here?  https://aviation.stackexchange.com/questions/56023/gravitational-power-vs-engine-power/56040#56040 and https://aviation.stackexchange.com/a/56040/34686  .  But weight * sink rate -- the approximation discussed in the answer-- is not the same as weight * inverse of glide ratio.

I was pointing out that weight x inverse glide ratio gave a fairly close answer for minimum power maintain altitude, and indeed, involves sink rate. But thanks, I will review these articles.


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Is Vy closer to Vbg with a variable pitch prop
My experience with fixed pitch props is that Vy is generally higher than Vbg due to engine efficiency issues. If I set my prop pitch so optimal engine rpm is at Vbg, will Vy = Vbg?

Vbg = V best power-off glide speed? Or are you thinking of a different definition?

Vbg = V  best lift to drag ratio, in other words, greatest aerodynamic efficiency.  At the higher speed the engine with fixed pitch prop likes, AOA would not be as optimal.  In other words, could I increase my climb angle and climb at 65 knots rather than 70 if my engine/prop had peak efficiency at 65 knots (Vbg).

@quiet flyer came across someone who published thrust data for Cessna 172P, actually close to 800 lbs with no forward motion, dropping to 600 lb near Vy, and around 400 lbs over 100 knots.  So, please go ahead and ask, this may really help to add to our knowledge.  These engineers really freak when we don't use the proper definitions, but thrust force at one given moment in time seems to be more applicable to compare with lift and drag.

I'm thinking that if you set your prop pitch so that the engine rpm was optimum (maximum amount of engine power generated at wide open throttle) at V bg, the optimum prop angle-of-attack would occur at some slightly different airspeed, so the maximum power output from the whole system would occur at some airspeed other than Vbg.  Would it be higher, or lower? Then on top of that consider the fact that that theoretical maximum climb rate ought to occur at Vmin sink, not Vbg.  So I think the answer to your question would be "probably not", but I don't know in which direction Vy would be shifted.

Too long for a comment:
My POH does not show Vbg, only the 'normal' V's listed below, and even then only some of these are indicated on the Airspeed Indicator with the Green, Yellow, Red color bands. How would you determine Vbg? Experimentally?
VA,  Maneuvering Speed is the maximum speed at which you many use abrupt control travel.
VFE,  Maximum Flap Extension Speed is the highest speed permissible with wing flaps in a prescribed extended position.
VNO,  Maximum Structural Cruising Speed is the speed that should not be exceeded except in smooth air, then only with caution.
VNe,  Never Exceed Speed is the speed limit that may not be exceeded at any time.
VS,  Stalling Speed or the minimum steady flight speed at which the airplane is controllable.
VSO,  Stalling Speed or the minimum steady flight speed at which the airplane is controllable in the landing configuration at the most forward center of gravity.
VX,  Best Angle-of-Climb Speed is the speed which results in the greatest gain of altitude in a given horizontal distance.
VY,  Best Rate-of-Climb Speed is the speed which results in the greatest gain in a given time.

Need answer from someone with variable pitch prop.  On your plane we'll go with Vy.


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Would it be possible to put V-22 Osprey engine/props on a PBY Catalina Flying Boat
The larger props of the Osprey may have greater efficiency over smaller props.  The high wing of the Catalina seems to have enough room for larger props.  The original PBY had a range of over 2000 miles and was amphibious.  Could this old design find use today with modern technology?

"The larger props of the Osprey may have greater efficiency over smaller props." why?

See question regarding power requirement comparison of helicopter vs plane.  We established that a higher aspect, slower (airfoil) prop is more efficient.  See also wing tip vortex cancelling concept of Vought V-173.  This work was stopped with the onset of the jet age, but may find application in the high cost of fuel age.

The engines on a PBY-5A Catalina are spaced only about 15' apart. The V-22 Osprey rotors (propellers) have a diameter of 36.1 feet and are much too large.
The engines would have to be moved more that 36' apart for the 18' blades to clear each other. The “high wing” of the Catalina is only about 10' from the water during water operations, and about 14' from the ground during ground operations, so that presents additional clearance problems.

To scale, putting V-22 props/engines on a Catalina, would look like this:

The drawing shows a prop diameter of 12' 1 ".  But more importantly, the distance between engines is approximately 15' which is less than the 18' radius of the Osprey prop blades. The props hitting the water would be trivial compared to the damage to the fuselage and the other engine.

Good point.... I have edited my answer in order to improve it. Thanks.

The engines would only need to be 18' and some change apart for the radius of the left prop to miss the right engine nacelle. They could be sync'd to interleave to avoid blade collision as they are on the Chinook (and they're linked on the Osprey itself, so the mechanicals are there). However, all the point for "why this won't work" are quite valid.

If the nacelles were mounted on the wings an adequate distance apart the rotors could be tilted to avoid striking the ground or water.  The V-22 can not land with rotors pitched down either.  Also, smaller rotors may be considered.  The Vought V-173 "flying pancake" also bears similarities.

@qq jkztd ok, please lower the nacelles to wing level and a little farther out to avoid striking the fuse.  Also, superimpose 25 foot rotors.  Thanks.


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Would a sharper leading edge increase lift and reduce drag at lower speeds?
A recent scratch built glider kept turning to the right even after painstaking efforts to match the wings perfectly.  Extensive addition of fin area to the tail failed to correct it.  Closer examination showed the left leading edge had been carved to a sharper point, angled more sharply than the curvature of the wing.
After carving the right LE to a point the glider flew perfectly straight.
This was not expected at the 5-7 mph the glider flies at, and the effect seemed to be stronger at lower speeds.
Could sharper leading edges help even at very low speeds?

Sand it off to match the other wing and see what the effect is. Personally, I don’t think this is causing what you describe.

@Greg Taylor actually i matched the other wing to it.  Both on the same plane made for a quite unexpected "wind tunnel" experiment.  Big relief pulling the extra lumber off the tail after it straightened out.  Made me think of the "old" B52 using spoilers instead of rudder and aileron.  Control surfaces are next.

The leading edge is round to allow a wider range of angles of attack. If the airfoil is operated at one angle of attack only, the leading edge can be sharp. See the leading edges of compressor blades or fowler flaps for example - the stator rsp. the wing ahead of them directs airflow such that the angle of attack variation on the airfoil is small, in case of flaps especially if the flap has only one extended position.
This means that your sharp leading edge works well because (by chance) it is exactly where the stagnation point in trimmed flight goes. Re-trim the aircraft for a different speed and the right-turning tendency will be reversed. Unfortunately, now you made both sides similar, so you cannot run this test directly.

Thanks so much for responding.  Lucky indeed as FF gliders usually fly at Vbg.  The right turn (much to my dismay) did not appear until velocity stabilized to trim speed as you have correctly diagnosed.  The "spike" may help generate forward lift by creating a less diffuse airstream near the leading edge, similar to what a slat does at higher AOA.  Wonder if LEs could be adjusted like inlet spikes to save fuel?  Will build more and test.

@RobertDiGiovanni: The F-18 does it (it's called "living wing") - the slats and flaps are continuously adjusted to the current angle of attack.

Other factors equal, a thinner ("sharper") leading edge will generally reduce drag in an airfoil.  There are tradeoffs, however; first and foremost in this case is the much more abrupt onset of stall as the sharper leading edge makes boundary layer detachment more abrupt.  In addition, the coefficient of lift will generally be lower, though peak L/D will often increase.
The combination of lower coefficient and sharper stall onset can make even a model cranky to fly -- in the case of a glider, a bad launch may cause a stall that devolves into a dive into the ground, even though a good launch might outperform the same model with a more conventional airfoil.  For competition hand-launch gliders, this is often tolerated and compensated by long practice at launching, in order to get the best possible glide performance.  For an R/C or piloted craft, on the other hand, this is a serious safety issue.

absolutely 100% with you on the launching technique.  This is why John Collins had a quarterback throw his indoor record "Suzanne" paper airplane over 226 feet.  Thanks!


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Where is the inclinometer "ball" in a knife edged turn?
Discussion of steep turns has lead to a question about where the inclinometer ball would be
if the plane rolled to 90 degrees.  Also, would there be any difference if the plane simply rolled to 90 degrees, rolled to 90 degrees and maintained altitude in a straight path with rudder, or rolled to 90 degrees and executed a "knife edged turn?
Does the ball become irrelevant at these attitudes?

As it was pointed out in the other thread, the ball just rolls around inside a curved glass tube.  It doesn't know what the angle of bank is, or anything other than what direction gravity, and/or centrifugal force is moving it.  Therefore:
At 90 deg AOB in coordinated flight at positive G, the ball would be centered.  The aircraft in this case would be in a descending spiral.
At 90 deg AOB in a "knife edge" straight pass where altitude is maintained with rudder, gravity will pull the ball towards the earth.  Same thing if you were able to make a knife edge turn holding altitude with rudder.

It isn't possible to be in coordinated flight at 90 degrees angle of bank, is it?

Why not?  You can be in coordinated flight inverted.  I didn't say you could do it while maintaining altitude.

Have you seen the video of Bob Hoover pouring iced tea while doing a barrel roll?

Yeah, I was thinking too narrowly.

It would be good to point out that in knife edge flight where rudder is being used to hold the nose up and there is no elevator input, the FUSELAGE is the wing and the wing is unloaded and doing nothing, just big fins sticking up and down.  In that case it's not really a turn.

I think if you are at 90 deg and still pulling, there is zero vertical lift vector so you are kind of in a ballistic spiral.

"At 90 deg AOB in coordinated flight at positive G, the ball would be centered. The aircraft in this case would be in a descending spiral." Mmmm, lemme run the numbers on that. If the bank angle not either 0 or 180 (meaning the lateral axis is not parallel to the horizon), then the only way the ball can be centered is if the aircraft is experiencing a lateral acceleration. For bank angles less than about, say, 75, this can be accomplished by making a coordinated turn. But steady coordinated flight at a bank angle of 90 degrees is impossible.

At 90 degrees of bank, and 0 degrees pitch (meaning the nose is pointed at the horizon), the situation is the most extreme: the only way to be in coordinated flight here is to be accelerating at 1 g downwards, which is to say, to be in freefall. So it's not correct to say that the aircraft would be in a descending spiral; it would have to be in a ballistic arc instead.

@Tanner Swett, I never said it would be steady state.  You are correct, it wouldn’t last very long…  One thing they teach you in IFR training is to recognize the dreaded “death spiral”.  With no horizon, (artificial or actual) if the aircraft starts a descending turn the pilot’s instinct is to pull back on the stick to correct.  As bank increases more backstick is applied and the spiral tightens as altitude lost increases.  The aircraft may transition though 90 deg AOB in coordinated flight.  If the condition isn’t recognized and corrected the aircraft will corkscrew right into the ground.

I also never said the pitch would be zero...


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How is rotor pitch best managed in helicopter autorotation?
In a recent stack exchange question writers were comparing autorotation to an aircraft engine out rotation prop when something rather startling occured to me:
In an engine out the airflow will continue to rotate the prop on the same direction as with power.  When helicopter loses power, WRT to rotor, it falls "backwards", creating force AGAINST the powered direction of rotation.
The thought is to go negative pitch and allow the rotor to build up angular momentum in the same direction as powered rotation, then switch to positive pitch to use that energy to flare and land.
Is this how it is done?

No, that is not how it is done: minimal collective blade pitch is not negative. The procedure is to lower collective immediately to the full down stop to minimise blade drag, upon which the helicopter will start to descend. The air upflow tilts the local lift vector forwards. Indeed, just like in a glider.
The bit that is not intuitively clear for fixed wingers is the reference frame of the airflow, which is the local blade velocity.
The inner bit of the rotor blade has a lower rotational speed than the tip. The up flow has a relatively larger influence on the lift vector tilt here, and  upflow through the rotor drives the inner part of the rotor blade forward, while the outer part still experiences decelerating torques. From Leishman:

It's similar to a fixed wing airplane really, except that it's happening in a circle.  It's done by reducing the rotor blade pitch to the minimum when power is initially lost, by lowering collective, to ensure the blades are able to maintain airspeed (in the form of rotor RPM), then adjusting pitch with collective as necessary to keep RPM in the green arc.
Take away the motive force, the engine and prop in the airplane, or the engine torque acting on the blade root in the helicopter, and you have to lower AOA to maintain flying speed, then adjust AOA to maintain the desired speed (or in the helicopter, RPM).  In autorotation, the flow switches from down through the rotor to up through the rotor because that's what it takes to maintain a flyable AOA.
The autorotating rotor is being driven around by the same forward thrust component of the lift vector that moves a gliding airplane forward.  The autorotating rotor is just like two gliders passing each other and who's wingtips stick together because the tips were made of Velcro, forcing them to glide in a circle centered around the joined wing tips.
Because they are rotary wings going round, the AOA is high near the root and lower near the tip because the blade speed increases going outboard.  Lift is perpendicular to AOA, so the thrust component of the lift vector is stronger near the middle and root than near the tip because the lift vector is canted forward, and the vertical lifting component of the lift vector is stronger near the tip because the lift vector is nearly straight up.
This is why they say most of the lift is coming from the outer part of the rotor disc and most of the driving force is coming from the inner and middle section of the disc (inner to mid span on each blade).
In an autorotation landing, inertia is used to convert forward energy of the blades to additional lift energy.  As you come out of the autorotation after the landing flare, you go back into "powered mode" you might say, by pulling pitch, changing the flow through the rotor from upward to downward as when the engine was running.  Except the "power" you are using isn't engine torque, it's just the inertia of the rotor blades.  You have a couple of seconds to use this inertia before the RPM decays and down you come.
It's a bit like settling a gliding airplane on the ground by pitching the nose up and trading inertial energy of the plane into temporary increased lift.

Turns out this is a fairly complex subject.  Apparently, if there is forward motion, the rotor can be "autogyroed" to generate lift.  But if it is purely vertical, the blade pitch would have to be really cranked down.  Seems that a forward motion is a big help with helicopters too.

Actually autogyros operate with a slightly positive blade pitch, one or two degrees, and they will autorotate straight down but the sink rate is too high for a good touchdown.  Except for a few homebuilt gyros with special long stroke gear that CAN land vertically without damage  https://www.youtube.com/watch?v=GlyR-aSEuig.

Plop!  Good gear.  There's a few things going on here.

Going to try an answer to bring a few power off lifting techniques to light.

Parachute:  uses drag force vector against gravity
Glider: uses drag force vector against gravity, perpendicular to a forward facing
 oblique surface, to generate  horizontal movement.  Wing uses movement to create lift.
Autogyro:  Uses air flow in plane of rotor to spin it (much like an anemometer).
 Generates forward motion by pitching down.  Spinning rotor creates lift.
100% Vertical descent with no forward component.  In order to maintain same rotor
 direction and RPM, at least part of the rotor must go negative.

Notice the descending rotor has its leading edge pitched UP in some diagrams, whereas
a gliders leading edge is always pointed down.  Simple pool hall physics tells us which way
the object will be deflected by the airstream.
There for, an autogyro, like a glider, will control its speed by pitching up or down, but a rotor in vertical descent with no sideways movement must alternate between generating RPM with
negative pitch and lift with positive.  In reality it can be set for a little bit of both, but the autogyro approach seems to be safer.


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Could the Boeing 757 replace the Boeing 737 Max 8/9/10?
The "Baby Boeing" 737 started life in the 1960s with roughly half the thrust and around 60% of the weight as the 757. In stages the 737 grew into a capacity squarely in the range of the 757. Originally, it prospered as a smaller regional jet. The Max 10 essentially will be a 757.
Is it possible the longer 757 airframe is a better match for the LEAP engines? Does the older, more forgiving AR 8 wing make it a safer aircraft?
Seems (forgive me) like comparing the graceful look of a thoroughbred, built for what it does best, compared to an overloaded old quarterhorse. Is the 757 worth another run?

References are from information provided in Wikipedia for Boeing 737 and 757 aircraft.

Have you ever tried to pull a load with a thoroughbred?

@Sean since airports are not standardized, with variations in altitude and temperature galore, these regional jets need to have, if anything, excess safety margins.  Excess power, oversized wings, good stability (especially for passenger carriers). Flying out of these "upside down wedding cakes"  on a finicky wing designed for optimal fuel efficiency just is not a good idea.  No horse is good if it lame, nor would I bring an economy car to the drag track.

Not for cheap it won’t.  The 757 went out of production in 2004.  Boeing has since dismantled the tooling for that airplane and it would cost a lot of money to get that set up again. Then all the subcontractors have to be brought up to speed as well.
Even if you could build 57s again - and there is a market for them - it would be getting increasingly outdated compared with modern airliners without a major upgrade to the standard avionics and systems.
In the end it’s probably much cheaper to address and fix the 737 MAX’s woes than to bring that airplane back.

Given the size of the existing 737 MAX fleet, there's no way it's not cheaper to address its problems. Scrapping the existing ~400 737 MAX aircraft and waiting around for the years it would take to develop a 757 MAX program is not a realistic alternative.

But when you spread that tooling cost across 5000 airplanes, ain't so bad.  Cheaper still would be a short wing root extension on the 737 allowing (causing) taller gear allowing tighter engine tuck.

Theoretically, yes! In practice, no
It all has to do with the typerating of pilots/crew and ground equipment. An aircraft like the 757 would be new to the fleet of many of the 737 Max operators. This means that they have to recertify pilots/cabin crew, purchase new ground handling equipment, train maintenance personel and invest in new stockparts.
The whole reason behind the debacle around the MCAS system has to do with the fact that by implementing MCAS the type rating pilots already had for the 737NG would still be valid on the 737 Max

Agreed on theoretical value and practical stopper regarding 757 MAX to replace 737 MAX. I don’t think MCAS, though, really is related to whether or not the type rating remains valid. The latter probably prevented an even more modern cockpit, but MCAS was as far as I know only required to meet certification criteria for stick force gradient at high angle of attack (i.e. even before discussing any type rating, the aircraft must meet this criteria before being certifiable). Control feel similarity to the NG didn’t matter too much, as far as I understand the issue.

@CptReynolds, without the MCAS system they had to change the design of the aircraft in order for it to be safe to fly. The MCAS is there to account for a design flaw, which is not necessarily a bad thing.

I don’t believe it would necessarily have been unsafe. It wouldn’t have met certification criteria, which is quite often a very similar thing but still subtly different.


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Is Engine up thrustline actually desirable for "speed stable" designs?
The much maligned Max 8 and the Buckeye Navy trainer.  What could they have in common?
Both would want static stability (trimmable speed stable approach) for landings.  Both would want maximum fuel efficiency for maximum range.
Engine up thrust would require less tail down trim for balancing forward CG.  Ideally, this could be optimized for cruise, so, at cruise airspeed and power, no tail down force is required, just like a plane with the CG back.  Much less drag.
The advantage is, once you drop out of cruise (less power), you now have a more staticly stable (read can be trimmed for one speed), making approaches to landing much easier.
Helpful for carriers and busy airports alike.
Is this such a bad design after all?

Bad for what purpose? Bad for any purpose, or bad for the purpose of "being a 737, just with brand new composites" (or some such)?

Well, if they get full power on/slow flight squared away I'd have to say awfully good for a 737!

Problem is if the thrust line effects are too strong it becomes a pain, especially if hand flying an approach and thrust changes require constant retrimming.  In any case the problem on the Max as I understand it isn't so much the thrust line effects but that the lift from the larger nacelles, moved forward, moved the aerodynamic neutral point forward, having the same effect as moving the CG limit aft.

Who said the Buckeye is much maligned?!!  I had a blast flying it, and so did everyone else I know.  As with every airplane you just need to learn the little idiosyncracies...

Didn't say it was.  Actually (since Navy aircraft designs are quite outstanding) I was trying to show this may be be a good idea if properly done.  All the diagrams I've ever seen never showed thrust angle as a means of improving cruise efficiency.  I also see a little 4FD in the Navy F35C.


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Can a plane fly without the vertical stabilizer?
What would happen if a plane (for the sake of the question let's say that the plane is an Airbus A380) lost its vertical stabilizer?

https://commons.wikimedia.org/wiki/File:Tail_of_a_conventional_aircraft.svg

The plane will probably crash.
The vertical stabilizer provides stability in yaw to conventional aircraft. Aircraft such as the B-2 manage to provide stability through computer control, and aircraft such as the Northrop flying wings are designed to fly without one. But if an aircraft designed to be stable using a vertical stabilizer loses that surface, it will be very difficult for pilots to stabilize manually with the remaining systems. While roll and differential thrust will both affect yaw, they will both be slower to react than a rudder, especially in a large aircraft like an A380. This can also damage the hydraulic systems, making it more difficult to control the remaining surfaces.
If experienced test pilots are at the controls (as in the B-52 incident below), or if the failure is anticipated and trained for, it's possible that the aircraft would be controllable enough to land safely. However, as the incidents below show, this kind of failure does not happen often, and can easily exceed the crew's ability to control the plane.
Examples where this has happened:
Japan Airlines Flight 123 lost most of its vertical stabilizer when the rear pressure bulkhead failed. Despite also losing hydraulic systems, the pilots managed to keep the plane in the air for a while, but ended up crashing into a mountain.
American Airlines Flight 587 lost its vertical stabilizer when the pilot's rudder inputs over-stressed the structure. It crashed shortly after.
In 1964, a B-52 lost most of its vertical stabilizer due to extreme turbulence. The pilots were able to extend air brakes on the wing tips to provide some stability. The Air Force sent up a chase plane to help guide the pilots, and provided engineering guidance from the ground. The crew were able to bring the plane back for a safe landing (see a video). There were at least three other instances where the vertical stabilizer failed on B-52's, all ending with the loss of the plane. This particular flight was conducted with test pilots to intentionally fly through turbulence, recording data to help understand the failures on the other aircraft. However, the flight encountered unexpectedly severe turbulence, leading to the separation of the vertical stabilizer.

Just as a point of reference, you don't actually need computers to stabilize without a horizontal stabilizer, the N9-M, a flying wing, didn't use one when it was developed in the early 40s (http://en.wikipedia.org/wiki/Northrop_N-9M)

I don't think that the Japan Airlines is a good example. In that case they lost all hydraulics,  so it's impossible to pinpoint the influence of the vertical stabilizer in the whole thing.

Jay Carr - I think you meant vertical stabilizer.

I wonder if airbus have designed this failure into their fly by wire system, if so the computers may be able to keep the airplane flying by using the ailerons as elvons.

@GdD I'm hoping that they have categorized loss of the vertical stabilizer as "extremely improbable," and thus not worth spending the money to program for.

@GreatTurtle Ooops...  Though technically the N9-M proves what I said just as much as what I didn't say... 0_0

Agreed, just pointing it out.

@GdD and fooot: Probably they cannot and should not (complexity is dangerous in multiple ways) include every failure into the fly-by-wire-systems. For this we have two pilots and Normal-, Alternate-, Direct- and Mechanical-Laws by Airbus and Normal- and Secondary-Mode by Boeing. Than apply hope for the best. Because cannot happen is wrong, everything will happen, just a matter of time.

As always, it depends. There are several things which provide directional stability:

The stabilizer, obviously,
Positive wing sweep,
Fins and
pusher propellers.

They have to work against the destabilizing parts:

The fuselage,
External tanks and stores,
Tractor propellers and
Forward-facing engine nacelles.

If the vertical is the only stabilizing part, losing it means to crash shortly after. Airplanes need to face forward to create sufficient lift. Momentary exceptions don't count: WW I fighter airplanes had very small vertical tails, and consequently low directional stability. Several German pilots of WW I perfected a technique where sudden rudder input would push the airplane beyond the stable regime of yaw angles, and the airplane would make a quick, full rotation around its vertical axis. This would scare the pilots of pursuing airplanes, because now the machine guns of their adversary would point at them. But there was no real possibility to aim, and the rotation speed was so high that very few shots would actually go in the direction of the pursuing aircraft.
The size of the vertical tail of multi-engined aircraft is driven by the need to counteract the yawing moment due to a failed engine. If all engines are running, the vertical tail needs only to be 1/3 as big. If low dutch roll stability is tolerated, it could be cut back even more. Therefore, if the other parts can contribute enough stability, a partial loss of the vertical surface can be survived. With a swept wing, the key is to fly in a flight regime where it helps the most, that is at low speed. In the case of the B-52H 61-023, which had this happen to it on January 10, 1964, additional help was provided by lowering the rear landing gear which contributed a small fin effect and by shifting the center of gravity forward by pumping fuel.

Also, if any more of the stabilizer had broken off, longitudinal stability would have been lost. Those guys were both lucky, and very skilled pilots.
61-023 was repaired and flew for more than 40 years after this incident. It was retired in 2008.

If I ask how the pusher propellors on the N9-M contributed to that aircrafts overall stability (in another questions), could you give a detailed answer?  I'd love to see that :).

@JayCarr: I could only refer you to NACA TN 2585: Propellers in Pitch and Yaw. This basically says that a propeller works like a wing of equal size. I would guess that the sweep had more effect, but the propellers should not be neglected, especially under power.

Do you mean wing sweep (I assume?). Maybe I'll ask about that instead.

@JayCarr: Yes, wing sweep. The magnitude of the effect depends on speed; at high angle of attack the stabilizing effect is bigger. The Horten flying wing gliders showed marginal directional stability at high speed.

@PeterKämpf shifting the center of gravity forward by pumping fuel what was the purpose of shifting cg forward even more by pumping fuel?  wouldn't the loss of a vstab already shift the cg forward?  or does a more-forward cg contribute to stability in general?

@erich: Yes, the loss of the vertical tail shifted the cg forward, but more forward shift was needed to regain positive lateral stability. Both longitudinal and lateral stability increase with a forward cg shift - think of it as if the plane is hinged around the cg and rotates around it in flight, both laterally and longitudinally.

How tractor propellers reduce static directional stability?

Is there any reference for that WW I pirouette?  Could any newer airplanes still do that?

@CamilleGoudeseune I have read about it but cannot give you a reference straight away. I will try to find one. The closest you come to this in a modern airplane is in a hammerhead turn - modern designs need to be much more directionally stable to achieve certification.

There are several planes that are designed to fly without vertical stabilizers (like the B-2 bomber, for example).  But they have very clever split flaps that make up for the lack of stability/control usually provided by a horizontal stabilizer/rudder setup.  Basically a split flap will open up and create more drag on the wing the aileron is placed on, pulling that side of the aircraft back.  Thus those aircraft use that system for horizontal stability.
If a plane is not designed to work without the verticle stabilizer, it would have problems if it lost it.  On craft with horizontal stabilizers, in general, if the plane gets a little bit sideways the airflow will push against the stabilizer and thus move the plane back in line.  If you lost that system, you'd have the find another way to compensate for yaw instability.
It is theoretically possible that you could use differential thrust between the engines to keep the plane in line.  Granted, the response would be very sluggish and difficult to control (especially on a big jet like the A380), but it's possible that it could be done given the right plane (one with quickly responding engines) and the right conditions (smooth air, basically).

Probably not. One sad example is American Airlines Flight 587.
If the plane loses just a small part of of the fin, it might be ok, but if it loses the fin completely two major things happen:

The plane becomes unstable in yaw. If the plane needed such a big surface behind its center of gravity (CG) to be stable there is almost no chance of flying without it.
There is a significant mass loss at a big distance from the CG. This will lead to a sudden shift of the CG towards the front of the plane which will make the plane dive, possibly beyond any chance of recovery.

Aggravating to the above is the fact that most likely the pilots wouldn't be aware of what actually happened and it would be very difficult for them to take proper action.

It can perform an emergency landing if there is no further damage at tail section such as elevators. On JAL 123 crash, vertical stabilizer was missing but that was not the reason why they crashed. JAL crash occurred because of failure on elevator controls.

After JAL 123 I would say yes, likely can fly meaning stay in the air, but the landing is another question.

Sure it will (aerodynamically speaking). As previously mentioned, it pretty much requires a digital flight control system though (aka 'fly by wire').
Lockheed Martin X-44 Manta is pretty much the F-22 without vertical flight control surfaces.

There's a big difference between designing a plane to not need a vertical stabilizer and losing one on a plane that is designed to need it, though. An A380 would be in big trouble if the vertical stabilizer separated... just like the A300 that crashed in Queens back in 2001. Incidentally, that was the last aviation incident on a flight operated by a major U.S. carrier to result in any passenger fatalities.

Why use a hypothetical aircraft for proof? The X-36 could do much the same and has flown already. However, thrust vectoring only helps as long as the engines are running and have their exhaust at the rear fuselage, and in an descent of an airliner they produce very little thrust. Also, certifying this would be a major undertaking. For now, this is not an option for commercial aircraft. Split ailerons might work, though (see B-2)

I am pretty sure a plane that has fly by wire will not crash.  Fly by wire controls the stability of a plane by adjusting the control surfaces.  When the flaps are out it will balance both sides to prevent yaw.  I see no reason why the failure of the vertical stabilizer cannot be compensated by controlling the flaps.  A pilot couldn't do that.  A computer yes.
Edit: I am not so sure anymore.  Loosing the whole vertical stabilizer is much worse than loosing just the rudder.  But I would say there still is hope.

A computer will do something only if it has been programmed to do so. If the eventuality of flying without the stabilizer was not foreseen and somehow coded and calibrated, the fly by wire may actually be counter-productive.

At the time the A320 came out I read that this very feature was being experimented.  An experimental software was able to translate standard commands to alternate flight controls, precisely replacing the rudder by the flaps.  The plane behaved properly, even though the response felt quite sluggish.  The problem was that this software was taking liberties with the very strict specification of what a flight computer was allowed to do.  But I would be surprised if after 25 years there had been no progress and this hadn't matured into a standard failover solution.

Rereading Emil's answer, there is a big difference between loosing the control of the rudder and loosing the whole vertical stabilizer.  I can imagine the second case is outside of the situations the flight computer is programmed for.

The question was about the second case that you mentioned, though: the complete loss of the vertical stabilizer.

Flight 587 was undergoing violent yaw rotations before the vertical stab snapped. Once it was lost, nothing stopped the plane's rotation and it just continued to yaw until entering a spin. If the vertical stabilizer had simply come off while the plane was in normal flight,  then who knows if it would have survived to land, but it probably would not have instantly spun out of control.
Other instances where planes crashed after losing the vertical stab are the XB-70, where it lost both its vertical stabs after a collision, and DHL Flight 611, also lost its vertical stab due to a collision. Flight 611 survived longer than the aircraft it struck, but broke up soon after due to stresses put on the plane from the violent yawing and dive. It should be noted that a modern airliner would lose its crucial hydraulics if the fin detaches, rendering all control surfaces unusable.

It is not hard to design a swept-wing aircraft to fly fine without a vertical tail.  For example a "Zagi" flying wing like this

  will still fly ok if the vertical tip fins are removed, though adverse yaw will be more pronounced, and the CG must be further forward than the aftmost CG that can be tolerated if a vertical tail or tip fins are present, or else the aircraft will yaw out of control and then "tumble" violently.  Similarly, most hang gliders don't have vertical tails.  However, if an airliner as per the original question could fly fine without a vertical tail, the vertical tail wouldn't be there.  The design has been been optimized for the tailed configuration and the complete removal of the entire vertical tail would probably have disastrous consequences.  Loss of a just a portion of the vertical tail could be tolerated as long as adverse conditions, such as failure of one or more engines, were avoided.  As far as the famous B-52 example goes, note that this aircraft has quite a lot of flat-sided fuselage area aft of the CG.

There are some airplanes from the 1920s that have no vertical stabilizer because the rectangular shape of their fuselage provided enough vertical stabillity. They were less manouverable but enough for the epoch. One example is the Fokker FII.

The Fokker FII had a vertical stabilizer. Some might argue that it was relatively small, but it was there nonetheless. https://flughafenbb.files.wordpress.com/2012/04/fokker-fii-d-742-bc3b6blingen_.jpg


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Is it allowed to listen to music while piloting a plane?
When I'm driving my car, I almost always put some music on. It's really relaxing for me.
I know that piloting a plane is nothing like driving a car, but it would be great, at least for me, to listen to some music while piloting.
So here is my question:

Is it allowed to listen to music while piloting a plane? Maybe with some device that mutes the music when someone is speaking in the radio?

Maybe it's just me, but this just feels wrong.  At least in helicopters, listening to the machine is another set of inputs which are as valuable as the instruments and what my behind is telling me.  I also wonder if relaxing is what''s needed and by how much you want to relax.  I find that being alert is a key part of the act of committing aviation.

@Simon There is, sadly, a difference between "legal" and "good idea".

@JayCarr I don't agree that it is sad that the difference exists in general. A state that does not allow individuals to define what a good or a bad idea is, is in its very fundamentals a totalitarian state. I cherish my freedom to have bad ideas and fail.

Do you guys think that music would take your attention from the flight instruments?

Wouldn't music be a way to cope with the boredom? http://aviation.stackexchange.com/questions/13422/how-do-long-haul-pilots-cope-with-boredom

@DeltaLima - I think you've mistaken an attempt at humility (and a general effort to not appear preachy), for some sort of political statement.  Good for you though, I would totally agree with your point, personally.  With the small caveat that sometimes safety measures are required to stop one persons "bad idea" from harming someone else.  What side of the fence this particular point falls on...  Eh, not sure.

A lot of my friends ski and ride motorcycles with headphones. I find it distracting.

@GabrielBrito To answer the question in your comment, we must assume that boredom is common?

@DeltaLima I’m not obliged to fly with you but I have little authority over whether I’m to be under your flight path.  Please don’t exercise your freedom to have bad ideas and fail whilst overhead.

Many aviation headsets either have an Aux-jack or a bluetooth connection, and listening to music is specifically mentioned as a feature.
(Link to headset)
(Link to adapter)
Some headsets even have a feature that detects radio-communications, and immediately silences the music.
The FAA has no problem with pilots listening to music provided its not distracting.   Individual company policies may differ.

If your plane has an old ADF receiver, you can also tune it to find nearby AM radio stations.   It doesn't sound very good, but you can hear 'em.

The Link to headset doesn't work. SprotsYS shows an ERR 404 page for the link

Fixed the link.

Yes you can listen to music while piloting a plane.  Many areas of the country have wide stretches where there is  little to no traffic, or air traffic control, to listen to. Other areas are quite congested and auto-muting music would be really annoying.  Boston, NYC, Albany areas as an example. West of Albany it gets quieter, until getting close to Rochester and then Buffalo.  I prefer no music myself, especially in areas new to me, so I can hear any directions from ATC that may come up.  My passengers (up to the 3 I can carry) are free to listen to music if they want to, tho I will ask all to listen and watch in the takeoff or landing environment.


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When do pilots engage the autopilot in airliners?
I may be wrong but I learned that in regular airliners, pilots always takeoff manually.
After taking-off and going to a positive rate of climb:

When do pilots engage the autopilot?
Is there a minimum altitude to engage it?

The minimum altitude for autopilot engagement is generally the acceleration height or around 400' AGL though this can vary between operators.
The question of when the autopilot is actually engaged is a more complicated question and is subject to pilot preference.  Personally, I generally hand flew until the first level off or until climbing and accelerating through 10,000 MSL.  Others may turn it on as soon as allowed and most fall somewhere in the spectrum between the two.

For me, it was a question of how close you were to your 6 monthly Sim check. The closer you were the more handling practice you gave yourself.
In general, on a normal line flight, I agree, around 400 feet on acceleration and as the BAe 146 had no Autothrust I generally hand flew most of the approach from decelerating, taking flap 18 and putting the gear down. It was my feeling that a human pilot coped better with configuration changes than the autopilot and with no autothrust I wanted to be completely in the loop and therefore, completely flying the aircraft rather than having the autopilot take care of 3 axes and juggling the thrust manually.